Question

28 Find mode and mean of the given distribution table
Class
1-10 11-20 21-30 31-40 41-50 51-60 61-70
Interval
Frequency
6 5 3 8 2 4 1​

Answer 1

Step-by-step explanation:

Interval 1-10 11-20 21-30 31-40 41-50 51-60 61-70

Interval 1-10 11-20 21-30 31-40 41-50 51-60 61-70Frequency 6 5 3 8 2 4 1

Interval 1-10 11-20 21-30 31-40 41-50 51-60 61-70Frequency 6 5 3 8 2 4 1mid value 5.5 15.5 25.5 35.5 45.5 55.5 65.5

Interval 1-10 11-20 21-30 31-40 41-50 51-60 61-70Frequency 6 5 3 8 2 4 1mid value 5.5 15.5 25.5 35.5 45.5 55.5 65.5ui -3 -2 -1 0 1 2 3

Interval 1-10 11-20 21-30 31-40 41-50 51-60 61-70Frequency 6 5 3 8 2 4 1mid value 5.5 15.5 25.5 35.5 45.5 55.5 65.5ui -3 -2 -1 0 1 2 3fiui -18 -10 -3 0 2 8 3

£fi= 29 £ fiui= -18 a= 35.5, h = 10

mean a+ £fiui/£fi ×h

= 35.5 + (-18/29) ×10

= 35.5-180/29

= 35.5-6.21

= 29.29

mean = 29.29 approximately

l= 30, f0= 3, f1= 8, f2= 2 , h= 10

mode = l +{ (f1-f0) /2f1-f0-f2)}× h

= 30 + { (8-3) /(16-3-2) }×10

= 30+ (5/11) ×10

= 30+ 50/11

= 30+4.545...

= 34.54..

mode = 34.54